# Thread: Challenging Limit Question!

1. ## Challenging Limit Question!

Ok I've got to evaluate the following limit:

But I've got no idea how the absolute values can be rewritten in this limit?

2. Originally Posted by Kataangel
Ok I've got to evaluate the following limit:

But I've got no idea how the absolute values can be rewritten in this limit?
Note that $\displaystyle \left|x-1\right|=\left\{\begin{array}{rl} x-1, & \text{ if }x\geq 1\\ {\color{red}-x+1}, & \text{ if }x<1\end{array}\right.$ and $\displaystyle \left|x+1\right|=\left\{\begin{array}{rl} {\color{red}x+1}, & \text{ if }x\geq -1\\ -x-1, & \text{ if }x<-1\end{array}\right.$

Thus, $\displaystyle \lim_{x\to 0}\frac{x}{\left|x-1\right|-\left|x+1\right|}=\lim_{x\to 0}\frac{x}{(-x+1)-(x+1)}=\lim_{x\to0}-\frac{x}{2x}$.

I'm sure you can take it from here.

3. Originally Posted by Chris L T521
Note that $\displaystyle \left|x-1\right|=\left\{\begin{array}{rl} x-1, & \text{ if }x\geq 1\\ {\color{red}-x-1}, & \text{ if }x<1\end{array}\right.$ and $\displaystyle \left|x+1\right|=\left\{\begin{array}{rl} {\color{red}x+1}, & \text{ if }x\geq -1\\ -x+1, & \text{ if }x<-1\end{array}\right.$

Thus, $\displaystyle \lim_{x\to 0}\frac{x}{\left|x-1\right|-\left|x+1\right|}=\lim_{x\to 0}\frac{x}{(-x-1)-(x-1)}=\lim_{x\to0}-\frac{x}{2x}$.

I'm sure you can take it from here.
Thanks very much for illustrating Chris, but I have no idea how you chose x-1 in the 2nd set of brackets in $\displaystyle \lim_{x\to 0}\frac{x}{(-x-1)-(x-1)}$???

And why the red ones in particular? I'm not sure which set of arbitrary values this limit should be taking? This would be easier with one sided limits.

4. Originally Posted by Kataangel
Thanks very much for illustrating Chris, but I have no idea how you chose x-1 in the 2nd set of brackets in $\displaystyle \lim_{x\to 0}\frac{x}{(-x-1)-(x-1)}$???

And why the red ones in particular? I'm not sure which set of arbitrary values this limit should be taking? This would be easier with one sided limits.
erm...woops. I took the wrong one from each group (and had a couple typos). Please note my correction.

I looked at the intervals of each piecewise term, and picked the ones that contained zero in the interval (since we're taking the limit as x approaches zero).

5. By the way, how do you know which of the greater than or less than signs you should put the equals sign under?

i.e. how can you tell which sign needs to have the 'or equal' with it because attaching the equals under greater than or less than are equally possible?

6. Originally Posted by Kataangel
By the way, how do you know which of the greater than or less than signs you should put the equals sign under?

i.e. how can you tell which sign needs to have the 'or equal' with it because attaching the equals under greater than or less than are equally possible?
It really doesn't matter which one. Its just a matter of preference, I guess.